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issue when running cmder.exe from a java application

I'm trying to run cmder.exe from a java application. I'm using the java ProcessBuilder class and it works great with cmd.exe, but never with cmder.exe.
This is my code:
command = "cd C:\\Users\\fxxx\\Documents\\plt 3.0\\git\\xxx-estatic";
ProcessBuilder builder = new ProcessBuilder("C:\\PLT 3.0\\abc\\cmder_mini\\cmder.exe", "/start", command);
I want to open the cmder command prompt and run a "cd" command. Cmder.exe is actually launched, but I get the "file name, directory name or volume label syntax is incorrect" error message and my command (cd...) is never executed.
Again, things work perfectly if I use cmd.exe.
Does someone know how to effectively run cmder.exe from a java application?
Thanks a lot.

RTFM!
Assuming that you use that cmder.exe, the doc states:
Cmder.exe Command Line Arguments
Argument Description
/C [user_root_path] Individual user Cmder root folder. Example: %userprofile%\cmder_config
/SINGLE Start Cmder is single mode.
/START [start_path] Folder path to start in.
/TASK [task_name] Task to start after launch.
That means that your code should be:
path = "cd C:\\Users\\fxxx\\Documents\\plt 3.0\\git\\xxx-estatic";
ProcessBuilder builder = new ProcessBuilder("C:\\PLT 3.0\\abc\\cmder_mini\\cmder.exe", "/start", path);

How to execute the shell script program file from the Java class

I have written a shell script file which extracts the files, Please see below.
File Name: unzip.sh
#/bin/sh
cd /home/zip;
UNZIPDIR=/home/unzip/;
for i in *.zip; do
unzip "$i" -d "$UNZIPDIR"
rm "$i";
done;
This shell script executes sucessfuly on putty,
$> ./zip.sh
As i wanted to execute this script from my java class while i have tried several ways to invoke/execute the shell script file but it's not executing. Please see below java code.
//First try
File executorDirectory = new File("/home/zip");
ProcessBuilder processBuilder = new ProcessBuilder("./unzip.sh");
processBuilder.directory(executorDirectory);
Process p = processBuilder.start();
p.waitFor()
//Second try
Process p = Runtime.getRuntime().exec("/home/zip/unzip.sh");

The problem is that you fail to account for the process' standard output/error (as mentioned by #Yazan in the comments). You need to .get{Output,Error}Stream() from the created process and read from them (even if it is only to discard it).
The real problem however is that you use an external shell script for something which is entirely doable in Java itself. See this page which gives an example of how to extract a zip file entirely with Java code; to delete a file, use Files.delete().

Executing commands with the ProcessBuilder in java

I want to run the following command from a java file command : "java hello < C:\iptest\input.txt > C:\outtest\name.txt "hello" will take input from "C:\iptest\input.txt" and will produce an output file at "C:\outtest\name.txt".
codes i have done
String command[]={"java","hello","< C:\\iptest\\input.txt >","C:\outtest\name.txt"};
ProcessBuilder pb=new ProcessBuilder(command);
pb.directory(new File("E:\"));
and now how to go forward i have no idea .Please Help!!

Here is a link to the ProcessBuilder definition on Oracle, which explains how to use this class. Typically, this is used for executing non-java processes, e.g. .BAT or .EXE processes, but I suppose you could use it to execute any process you wish.
http://docs.oracle.com/javase/7/docs/api/java/lang/ProcessBuilder.html

Change PWD of linux from JSP

I need to execute linux commands from JSP.
It is working fine.
But i need to start some sh file in a particular directory in linux through JSP. say /home/username/something/start.sh
try{
String command= "cd /home/username/something";
Runtime.getRuntime().exec(command);
Runtime.getRuntime().exec("./start.sh")
out.println("Child");
}
catch(Exception e)
{ out.println("Error");
}
It says FIle or Directory not found.
I tried Runtime.getRuntime().exec("pwd"), It is showing something like "java.lang.UNIXProcess#fc9d2b" !! :O
I need to change the pwd and execute some commands through jsp. How can i do that??
Any help would be appreciated.

You shouldn't (and actually, it seems you can't) set a working directory like that. Each Process object given by Runtime.exec() will have its own working directory.
As answered in How to use “cd” command using java runtime?, you should be using the three argument version of Runtime.exec(), in which you provide a File that will be the working directory. From its javadoc:
Executes the specified command and arguments in a separate process with the specified environment and working directory.
Or even better, use ProcessBuilder along with ProcessBuilder.directory() instead:
ProcessBuilder pb = new ProcessBuilder("start.sh");
pb.directory(new File("/home/username/something"));
Process p = pb.start();

How to execute unix commands through Windows/cygwin using Java

I am trying to accomplish two things:
I am running cygwin on Windows7 to execute my unix shell commands and I need to automate the process by writing a Java app. I already know how to use the windows shell through Java using the 'Process class' and Runtime.getRuntime().exec("cmd /c dir"). I need to be able to do the same with unix commands: i.e.: ls -la and so forth. What should I look into?
Is there a way to remember a shell's state?
explanation: when I use: Runtime.getRuntime().exec("cmd /c dir"), I always get a listing of my home directory. If I do Runtime.getRuntime().exec("cmd /c cd <some-folder>") and then do Runtime.getRuntime().exec("cmd /c dir") again, I will still get the listing of my home folder. Is there a way to tell the process to remember its state, like a regular shell would?
It seems that the bash command line proposed by Paŭlo does not work:
C:\cygwin\bin>bash -c ls -la
-la: ls: command not found
I am having trouble figuring out the technicalities.
This is my code:
p = Runtime.getRuntime().exec("C:\\cygwin\\bin\\bash.exe -c ls -la");
reader2 = new BufferedReader(new InputStreamReader(p.getInputStream()));
line = reader2.readLine();
line ends up having a null value.
I added this to my .bash_profile:
#BASH
export BASH_HOME=/cygdrive/c/cygwin
export PATH=$BASH_HOME/bin:$PATH
I added the following as well:
System Properties -> advanced -> Environment variables -> user variebales -> variable: BASH, value: c:\cygwin\bin
Still nothing...
However, if I execute this instead, it works!
p = Runtime.getRuntime().exec("c:\\cygwin\\bin\\ls -la ~/\"Eclipse_Workspace/RenameScript/files copy\"");

1. Calling unix commands:
You simply need to call your unix shell (e.g. the bash delivered with cygwin) instead of cmd.
bash -c "ls -la"
should do. Of course, if your command is an external program, you could simply call it directly:
ls -la
When starting this from Java, it is best to use the variant which takes a string array, as then
you don't have Java let it parse to see where the arguments start and stop:
Process p =
Runtime.getRuntime().exec(new String[]{"C:\\cygwin\\bin\\bash.exe",
"-c", "ls -la"},
new String[]{"PATH=/cygdrive/c/cygwin/bin"});
The error message in your example (ls: command not found) seems to show that your bash can't find the ls command. Maybe you need to put it into the PATH variable (see above for a way to do this from Java).
Maybe instead of /cygdrive/c/cygwin/bin, the right directory name would be /usr/bin.
(Everything is a bit complicated here by having to bridge between Unix and Windows
conventions everywhere.)
The simple ls command can be called like this:
Process p = Runtime.getRuntime().exec(new String[]{"C:\\cygwin\\bin\\ls.exe", "-la"});
2. Invoking multiple commands:
There are basically two ways of invoking multiple commands in one shell:
passing them all at once to the shell; or
passing them interactively to the shell.
For the first way, simply give multiple commands as argument to the -c option, separated by ; or \n (a newline), like this:
bash -c "cd /bin/ ; ls -la"
or from Java (adapting the example above):
Process p =
Runtime.getRuntime().exec(new String[]{"C:\\cygwin\\bin\\bash.exe",
"-c", "cd /bin/; ls -la"},
new String[]{"PATH=/cygdrive/c/cygwin/bin"});
Here the shell will parse the command line as, and execute it as a script. If it contains multiple commands, they will all be executed, if the shell does not somehow exit before for some reason (like an exit command). (I'm not sure if the Windows cmd does work in a similar way. Please test and report.)
Instead of passing the bash (or cmd or whatever shell you are using) the commands on the
command line, you can pass them via the Process' input stream.
A shell started in "input mode" (e.g. one which got neither the -c option nor a shell script file argument) will read input from the stream, and interpret the first line as a command (or several ones).
Then it will execute this command. The command itself might read more input from the stream, if it wants.
Then the shell will read the next line, interpret it as a command, and execute.
(In some cases the shell has to read more than one line, for example for long strings or composed commands like if or loops.)
This will go on until either the end of the stream (e.g. stream.close() at your side) or executing an explicit exit command (or some other reasons to exit).
Here would be an example for this:
Process p = Runtime.getRuntime().exec(new String[]{"C:\\cygwin\\bin\\bash.exe", "-s"});
InputStream outStream = p.getInputStream(); // normal output of the shell
InputStream errStream = p.getInputStream(); // error output of the shell
// TODO: start separate threads to read these streams
PrintStream ps = new PrintStream(p.getOutputStream());
ps.println("cd /bin/");
ps.println("ls -la");
ps.println("exit");
ps.close();

You do not need cygwin here. There are several pure Java libraries implementing SSH protocol. Use them. BTW they will solve your second problem. You will open session and execute command withing the same session, so the shell state will be preserved automatically.
One example would be JSch.